Metal + Acid Displacement

Metal + Acid Displacement

Activity Series of Metals

Activity Series of Metals

• metals higher in series react with compounds of those below

• metals become less reactive to water top to bottom

• metals become less able to displace H2 from acids top to bottom

Potassium + Water

Activity Series of Metals

Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)

Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s)

Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g)

Zn(s) + 2HBr(aq) ZnBr2(aq) + H2(g)

Metal + Metal Salt Displacement

Which of the following reactions does NOT happen?

100

0

1300% 0% 0%0%0%

1. Cu(s)+H2SO4(aq)CuSO4(aq)+H2(g)

2. 2HNO3(aq)+2K(s)2KNO3(aq)+H2(g)

3. FeCl2(aq)+Zn(s)ZnCl2(s)+Fe(s)

4. Ca(s)+2H2O(l)Ca(OH)2(aq)+H2(g)

5. Cu(s)+2AgNO3(aq)2Ag(s)+Cu(NO3)2(aq)

Solution

A homogeneous mixture of two or more substances comprising the solvent which is the majority of the mixture and one or more solutes which are the smaller fraction.

Concentration – how much solute is present in a given amount of solution

Cola DrinksSolvent

• water

Solutes

• carbon dioxide (gas)

• sweetener (solid)

• phosphoric acid (liquid)

• caramel color (solid)

Molarity – a measure of concentration

The number of moles of solute per liter of solution.

molarity M

moles of soluteM =

liters of solution

units molar = mol/L = M

Preparation of 1.00 L of 0.0100 M

KMnO4 solution from solid

Which value do you NOT need to determine the molarity of a solution?

100

0

130

Mas

s of s

olute

Mola

r mas

s of s

olute

Volum

e of s

olve

nt a

dded

Total v

olum

e of s

olutio

n

0% 0%0%0%

1. Mass of solute

2. Molar mass of solute

3. Volume of solvent added

4. Total volume of solution

Solution Preparation by Dilution

Dilution

• Molarity (mol/L) × Volume (L) = Moles

• If we take a sample (say 25.0 mL) from a solution (say 0.372 M) and add extra water (say to a total volume of 500. mL) the moles of solute are unchanged

• Thus M1V1 = moles = M2V2

• 0.372 M × 25.0 mL = M2 × 500. mL

• M2 = 0.0186 M

How many L of conc HNO3 (16.0 M) are needed to prepare 0.500 L of 0.250 M nitric acid?

100

0

130

32.0

L

16.0

L

0.50

0 L

0.25

0 L

0.00

781

L

0% 0% 0%0%0%

1. 32.0 L2. 16.0 L3. 0.500 L4. 0.250 L5. 0.00781 L

Stoichiometric Relationships

Titrationsat equivalence mol H+ = mol OH-

EXAMPLE: A sample of lye, sodium hydroxide, is neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?

2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O

(25.0 mL NaOH) #mL H2SO4 =

(0.400 mol NaOH)

(1 L NaOH)

(1 L)

(1000 mL)

(1 mol H2SO4)

(2 mol NaOH)

(1 L H2SO4)

(0.200 mol H2SO4)

= 25.0 mL H2SO4

(1000 mL)

(1 L)

Ion Concentrations

• 0.100 M NaCl

• NaCl(aq) Na+(aq) + Cl-

(aq)

• Is 0.100 M in both Na+ and Cl-

• 0.100 M Al2(SO4)3

• Al2(SO4)3(aq) 2 Al3+(aq) + 3 SO4

2-(aq)

• Is 0.200 M in Al3+ and 0.300 M in SO42-

• Consider the NaOH + H2SO4 reaction. What are the final concentrations of Na+ and SO4

2-?

• Stoichiometric reaction, can use either reactant to determine moles of product

• 25.0 mL NaOH × 0.400 mol NaOH × 1 L 1 L 1000 mL0.0100 mol NaOH × 1 mol Na2SO4 = 0.0050 mol Na2SO4

2 mol NaOH

Final Ion Concentrations

• Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 0.0500 L

• [Na2SO4] = 0.00500 mol = 0.100 M 0.0500 L

• [Na+] = 2 × 0.100 M = 0.200 M

• [SO42-] = 0.100 M

Which solution has the highest concentration of SO4

2-?

100

0

130

0.20

M C

uSO4

0.15

M N

a2SO

4

0.07

0 M

Fe2

(SO4)

3

0.10

M C

e(SO4)

2

0% 0%0%0%

1. 0.20 M CuSO4

2. 0.15 M Na2SO4

3. 0.070 M Fe2(SO4)3

4. 0.10 M Ce(SO4)2